Problem: Divide the following complex numbers. $ \dfrac{7+17i}{2+3i}$
Explanation: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${2-3i}$ $ \dfrac{7+17i}{2+3i} = \dfrac{7+17i}{2+3i} \cdot \dfrac{{2-3i}}{{2-3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(7+17i) \cdot (2-3i)} {(2+3i) \cdot (2-3i)} = \dfrac{(7+17i) \cdot (2-3i)} {2^2 - (3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(7+17i) \cdot (2-3i)} {(2)^2 - (3i)^2} = $ $ \dfrac{(7+17i) \cdot (2-3i)} {4 + 9} = $ $ \dfrac{(7+17i) \cdot (2-3i)} {13} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({7+17i}) \cdot ({2-3i})} {13} = $ $ \dfrac{{7} \cdot {2} + {17} \cdot {2 i} + {7} \cdot {-3 i} + {17} \cdot {-3 i^2}} {13} $ Evaluate each product of two numbers. $ \dfrac{14 + 34i - 21i - 51 i^2} {13} $ Finally, simplify the fraction. $ \dfrac{14 + 34i - 21i + 51} {13} = \dfrac{65 + 13i} {13} = 5+i $